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Math Help - negative exponent elimination

  1. #1
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    negative exponent elimination

    Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

    (x^-3)(x^2)
    ------------
    (x^-7)

    becoming

    (x^2)(x^7)
    -----------
    (x^3)

    reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
    x^5
    ----
    x^-4
    and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
    (x^4)(x^-6)
    ------------
    (x^-5)(x^-2)

    thanks once again
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  2. #2
    YOurBoYFrOmDaHoOd
    Guest
    [quote=ScottPaulsen1;68439]Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

    (x^-3)(x^2)
    ------------
    (x^-7)

    becoming

    (x^2)(x^7)
    -----------
    (x^3)

    reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
    x^5
    ----
    x^-4
    and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
    (x^4)(x^-6)
    ------------
    (x^-5)(x^-2)

    Ok, man you need to go to this site it has everything you need

    Studyit: Studyit - for NCEA students its really good Ese
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ScottPaulsen1 View Post
    Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

    (x^-3)(x^2)
    ------------
    (x^-7)

    becoming

    (x^2)(x^7)
    -----------
    (x^3)

    reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
    x^5
    ----
    x^-4
    and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
    (x^4)(x^-6)
    ------------
    (x^-5)(x^-2)

    thanks once again
    Remember we can always take a number x and put a 1 under it: x = \frac{x}{1}

    So
    x^5x^{-8} = x^5 \cdot \frac{x^{-8}}{1} = \frac{x^5}{x^8}

    In the case of \frac{x^5}{x^{-4}} = \frac{x^5x^4}{1} = x^5x^4

    Finally:
    \frac{x^4x^{-6}}{x^{-5}x^{-2}} = \frac{x^4x^5x^2}{x^6}

    -Dan
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